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Evaluate nth Roots and Use Rational Exponents

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1 Evaluate nth Roots and Use Rational Exponents
Chapter 6.1

2 Nth Roots In an earlier chapter, we defined the square root of a number as follows If π‘₯ 2 =π‘Ž, then π‘₯=Β± π‘Ž When solving for a square, we can ask, β€œWhat number multiplies itself to produce a?” For example, if π‘₯ 2 =4, then x is the number that multiplies itself to produce 4 It turns out that there are actually two such numbers: 2 and βˆ’2, because 2 2 =4 and βˆ’2 2 =βˆ’2β‹…βˆ’2=4

3 Nth Roots We can extend this idea further by asking of some number a, β€œWhat multiplies itself three times to produce a?” We can express this as π‘₯ 3 =π‘Ž For example, what number multiplies itself three times to produce 8? This is the solution to the equation π‘₯ 3 =8 Since 2β‹…2β‹…2= 2 3 =8, then π‘₯=2; note that there is only one solution We can define the cube root of a number as If π‘₯ 3 =π‘Ž, then π‘₯= 3 π‘Ž

4 Nth Roots We can define the cube root of a number as If π‘₯ 3 =π‘Ž, then π‘₯= 3 π‘Ž Notice that, unlike a square root, a cube root has only one answer Also unlike a square root, the cube root of any real number is a real number Recall that if π‘₯ 2 =βˆ’1, then x is not a real number; generally, if π‘₯ 2 equals any negative number, the result is imaginary But if π‘₯ 3 =βˆ’8, then π‘₯= 3 βˆ’8 =βˆ’2 because βˆ’2β‹…βˆ’2β‹…βˆ’2=βˆ’8

5 Nth Roots Can we extend this further to 4th roots?
If π‘₯ 4 =π‘Ž, then x is the number that multiplies itself four times to produce a For example, if π‘₯ 4 =81, then π‘₯=3 because 3β‹…3β‹…3β‹…3= 3β‹…3 β‹… 3β‹…3 =9β‹…9=81 But x can also be βˆ’3 because βˆ’3β‹…βˆ’3 β‹… βˆ’3β‹…βˆ’3 =9β‹…9=81 So we define a fourth root as If π‘₯ 4 =π‘Ž, then π‘₯=Β± 4 π‘Ž

6 Nth Roots So we define a fourth root as If π‘₯ 4 =π‘Ž, then π‘₯=Β± 4 π‘Ž
As will square roots, the fourth root of a negative number is imaginary For example, π‘₯ 4 =βˆ’16 has no real solution because there is no real number that multiplies itself 4 times to produce βˆ’16 We will only be concerned with real numbers in this chapter, so we will say that this has no real solution

7 Nth Roots For the notation 𝑛 π‘Ž we would say β€œthe nth root of a”
The whole number n is called the index of the radical and a is again called the radicand Only two roots have special names: square root and cube root Also, the index for a square root is not written; that is 2 π‘Ž = π‘Ž In defining nth roots, we must differentiate between even and odd roots

8 n is an even whole number
Nth Roots Suppose that n is a whole number greater than 1, and that a is a real number such that π‘₯ 𝑛 =π‘Ž n is an even whole number n is an odd whole number If π‘Ž<0, there are no real nth roots If π‘Ž<0, then π‘₯= 𝑛 π‘Ž (1 solution, negative) If π‘Ž=0, then π‘₯= 𝑛 0 =0 (1 solution) If π‘Ž>0, then π‘₯=Β± 𝑛 π‘Ž (2 solutions) If π‘Ž>0, then π‘₯= 𝑛 π‘Ž (1 solution, positive)

9 Guided Practice Use the definition of nth roots and your calculator to find x. π‘₯ 5 =1024 π‘₯ 4 =16 π‘₯ 3 =βˆ’125 π‘₯ 6 =βˆ’1 π‘₯ 5 =2

10 Guided Practice Use the definition of nth roots and your calculator to find x. π‘₯ 5 =1024; π‘₯= =4 because 4β‹…4β‹…4β‹…4β‹…4=1024 π‘₯ 4 =16; π‘₯= 4 16 =2 because 2β‹…2β‹…2β‹…2=16 π‘₯ 3 =βˆ’125; π‘₯= 3 βˆ’125 =βˆ’5 because βˆ’5β‹…βˆ’5β‹…βˆ’5=βˆ’125 π‘₯ 6 =βˆ’1; π‘₯= 6 βˆ’1 , but since the radicand is negative, there is no real solution π‘₯ 5 =2; π‘₯= 5 2 β‰ˆ because (1.1487)(1.1487)(1.1487)(1.1487)(1.1487)β‰ˆ2

11 Rational Exponents Up to now we have only used integer exponents
Recall that π‘Ž 𝑛 =π‘Žβ‹…π‘Žβ‹…β€¦β‹…π‘Ž, where there are n factors of a Also, π‘Ž βˆ’π‘› = 1 π‘Ž 𝑛 = 1 π‘Ž β‹… 1 π‘Ž ⋅…⋅ 1 π‘Ž , where there are n factors of 1 π‘Ž We can extend the meaning of an exponent to rational numbers (that is, to fractions) To do this we must recall one of the properties of exponents: The power property says that π‘Ž 𝑛 π‘š = π‘Ž π‘›π‘š (multiply the exponents) Also recall that 𝑛⋅ 1 𝑛 =1 (inverse property of multiplication)

12 Rational Exponents How can we interpret the equation π‘₯= 25 1 2 ?
Suppose that we raise both sides to the power 2: π‘₯ 2 = By the power property of exponents we get π‘₯ 2 = β‹…2 By the inverse property of multiplication, π‘₯ 2 = 25 1 =25 But by the definition of square root, π‘₯= 25 =5 (we don’t need both positive and negative here) Now, since π‘₯= and π‘₯= 25 , then 25 = =5

13 Rational Exponents More generally, if π‘₯= π‘Ž 1 𝑛 , then π‘₯ 𝑛 = π‘Ž 1 𝑛 𝑛 = π‘Ž 1 𝑛 ⋅𝑛 = π‘Ž 1 =π‘Ž Since π‘₯ 𝑛 =π‘Ž, then by the definition of nth roots π‘₯= 𝑛 π‘Ž Now, π‘₯= π‘Ž 1 𝑛 and π‘₯= 𝑛 π‘Ž , so we conclude that π‘Ž 1 𝑛 = 𝑛 π‘Ž Rational exponents of the form 1 𝑛 are the same thing as nth roots

14 Rational Exponents This means that we can rewrite nth roots of a number as the number raised to the 1 𝑛 power, and vice verse For example, 4 7 = , 3 9 = , and 2 = Remember that we don’t write the index for the square root Also, = 6 , = 5 3 , and = 4 10

15 Guided Practice Rewrite the expression using rational exponent notation. 3 5 = 15 = 8 6 = 𝑛 𝑝 =

16 Guided Practice Rewrite the expression using rational exponent notation. 3 5 = 15 = 8 6 = 𝑛 𝑝 = 𝑝 1 𝑛

17 Guided Practice Rewrite the expression using radical notation. 4 1 3 =
= = = π‘₯ 1 𝑛 =

18 Guided Practice Rewrite the expression using radical notation.
= 3 4 = 5 11 = 3 π‘₯ 1 𝑛 = 𝑛 π‘₯

19 Rational Exponents With Denominator Other Than 1
Next we must interpret the meaning of π‘š 𝑛 as an exponent, where m is not 1 We can rewrite π‘š 𝑛 as π‘šβ‹… 1 𝑛 or as 1 𝑛 β‹…π‘š (commutative property) Now, by the power property of exponents π‘Ž 1 𝑛 π‘š = π‘Ž 1 𝑛 β‹…π‘š = π‘Ž π‘š 𝑛 But we also have that π‘Ž 1 𝑛 = 𝑛 π‘Ž , so π‘Ž 1 𝑛 π‘š = 𝑛 π‘Ž π‘š = π‘Ž π‘š 𝑛

20 Rational Exponents With Denominator Other Than 1
Since multiplication is commutative, we have π‘Ž π‘š 1 𝑛 = π‘Ž π‘šβ‹… 1 𝑛 = π‘Ž π‘š 𝑛 But this can also be written π‘Ž π‘š 1 𝑛 = 𝑛 π‘Ž π‘š = π‘Ž π‘š 𝑛 In general, π‘Ž π‘š 𝑛 = 𝑛 π‘Ž π‘š = 𝑛 π‘Ž π‘š Note that the denominator becomes the index and the numerator is an exponent in the radical or outside the radical (both are correct)

21 Rational Exponents With Denominator Other Than 1
Example: rewrite the expression with rational exponents as a radical expression = = = 3 16 β‰ˆ2.520 = = = 4 32 β‰ˆ2.378 βˆ’ = 5 βˆ’3 3 = 5 βˆ’3 3 = 5 βˆ’27 β‰ˆβˆ’1.933

22 Guided Practice Rewrite the expression as a radical, then evaluate using a calculator. = = βˆ’ = βˆ’ =

23 Guided Practice Rewrite the expression as a radical, then evaluate using a calculator. = = 3 2 =9 = = 2 3 =8 βˆ’ = 3 βˆ’ = βˆ’4 4 =256 βˆ’ = 5 βˆ’9 3 β‰ˆβˆ’3.737

24 Negative Rational Exponents
We defined negative exponents to mean π‘₯ βˆ’π‘› = π‘₯ βˆ’π‘› 1 = 1 π‘₯ 𝑛 That is, you can move the numerator to the denominator and change the sign of the exponent We can understand negative rational exponents using this definition π‘₯ βˆ’ π‘š 𝑛 = 1 π‘₯ π‘š 𝑛 = 1 𝑛 π‘₯ π‘š = 1 𝑛 π‘₯ π‘š , π‘₯β‰ 0

25 Negative Rational Exponents
Some examples 8 βˆ’ = = = = 1 16 βˆ’64 βˆ’ = 1 βˆ’ = βˆ’ = 1 βˆ’4 2 = 1 16 4 βˆ’ = = = β‰ˆ0.574

26 Guided Practice Rewrite the expression as a radical, then evaluate using a calculator. βˆ’30 βˆ’ = 9 βˆ’ = 243 βˆ’ =

27 Guided Practice Rewrite the expression as a radical, then evaluate using a calculator. βˆ’30 βˆ’ = βˆ’ = β‰ˆ0.104 9 βˆ’ = = = β‰ˆ 243 βˆ’ = = = 1 9

28 Solving Radical Equations
We were able to use the square root property to solve equations We can also use the nth root definition to solve equations In general, if π‘₯ 𝑛 =π‘Ž, then π‘₯= 𝑛 π‘Ž Some examples 4 π‘₯ 5 =128⟹ π‘₯ 5 =32⟹π‘₯= 5 32 =2 π‘₯βˆ’3 4 =21⟹π‘₯βˆ’3=Β± 4 21 ⟹3Β± π‘₯= β‰ˆ5.141, π‘₯=3βˆ’ 4 21 β‰ˆ0.859

29 Guided Practice Solve the equation. Round the result to three decimal places when appropriate. π‘₯ 3 =64 1 2 π‘₯ 5 =512 π‘₯βˆ’2 3 =βˆ’14

30 Guided Practice Solve the equation. Round the result to three decimal places when appropriate. π‘₯ 3 =64⟹π‘₯= 3 64 =4 1 2 π‘₯ 5 =512⟹ π‘₯ 5 =1024⟹π‘₯= =4 π‘₯βˆ’2 3 =βˆ’14⟹π‘₯βˆ’2= 3 βˆ’14 ⟹π‘₯=2+ 3 βˆ’14 β‰ˆβˆ’0.410

31 Exercise 6.1 Handout


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